# Conserved quantities with charge and Killing's vectors

I'm trying to solve the following problem:

A particle with electric charge e moves with 4-velocity $$U_{\alpha}$$ in a spacetime with metric $$g_{αβ}$$ in the presence of a vector potential $$A_µ$$. The equation describing this particle’s motion can be written $$U^{\beta} \nabla_{\beta} U_{\alpha} = e(\nabla_{\alpha} A_{\beta} - \nabla_{\beta} A_{\alpha})U^{\beta}$$.

The spacetime admits a Killing vector field $$ξ^α$$ such that $$\cal L_{\xi} g_{αβ} = 0,$$ $$\mathcal{L}_{\xi} A_α = 0.$$

Show that the quantity $$(U_α + eA_α)ξ^α$$ is constant along the worldline of the particle.

With all the derivatives introduced in GR, I'm slightly confused about what constant exactly means in this case. Nevertheless, I tried taking the Lie derivative of the above quantity and used the fact that $$\cal L_{\xi} A_{\alpha} = 0$$ to arrive at the following:

$$L_{\vec {U}} (U_{\alpha} + eA_{\alpha})ξ^{\alpha} = \xi^{\alpha} e U^{\gamma} \nabla_{\alpha} A_{\gamma} + U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha} + U^{\gamma} e A_{\alpha} \nabla_{\gamma} \xi^{\alpha},$$

using $$\cal L_{\xi} A_{\alpha} = 0$$, I can convert $$\xi^{\alpha} \nabla_{\alpha} A_{\gamma} = - A_{\alpha} \nabla_{\gamma} \xi^{\alpha}$$ to get

$$L_{\vec {U}} (U_{\alpha} + eA_{\alpha})ξ^{\alpha} = U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha}$$

which is clearly not zero. I haven't used the fact that $$\cal L_{\xi} g_{αβ} = 0$$ but I don't know how to apply it here. Also, I know this question was treated in Conserved quantities from Killing vectors in the presence of electric charge but there they use the action and we haven't talked much about that in the course, so I think it is not necessarily meant to be solved with it. Any ideas?

What you've written all looks correct, and you were very close to the result. You were along the right lines with needing $$\mathcal{L}_{\xi}g_{\mu\nu}=0$$. Your remaining term can be written as $$U^{\gamma} U_{\alpha} \nabla_{\gamma} \xi^{\alpha} = U^{\gamma} U^{\alpha} \nabla_{\gamma} \xi_{\alpha} \, ,$$ which comes from metric compatibility. The Lie derivative of the metric is $$\mathcal{L}_{\xi}g_{\mu\nu}= 2 \nabla_{(\mu} \xi_{\nu)}$$, which vanishes for Killing vector fields. Hopefully with these two equations you can now see why the remaining term is zero.
• Sorry, I still don't see it. The fact that $\mathcal{L}_{\xi}g_{\mu\nu}= 2 \nabla_{(\mu} \xi_{\nu)}$ tells me that $\nabla_{\alpha} \xi_{\beta} + \nabla_{\beta} \xi_{\alpha} = 0$, but I need to somehow show that $\nabla_{\gamma} \xi_{\gamma}$. What should I do? Also, why should I take the Lie derivative and not, for example, the derivative corresponding to Parallel transport? Mar 22 at 20:57
• It tells you that the tensor $T_{\mu \nu} := \nabla_\mu \xi_\nu$ is antisymmetric... There are lots of intuitive explanations of what the Lie derivative is. More or less, $\mathcal{L}_X T$ measures how a tensor $T$ changes along the flow generated by vector field $X$, so in your case you want to know how $Q$ changes along the flow of the particle, which is generated by (ie its tangent vector is) its four-velocity. Mar 22 at 21:35
• @SamuelJaramillo This was already answered by the other commented, but you should be able to see why the contraction $V^{\mu} V^{\nu} T_{\mu \nu}$ selects only the symmetric part of the tensor $T_{\mu \nu}$, which in this case is $\nabla_{\mu} \xi_{\nu}$. Mar 22 at 22:03